A child is in danger of drowning in a river and is being carried downstream by a current with a speed of 2.25km/h. The child is 0.645 km from the shore and 0.775 km upstream of a boat landing when a rescue boat sets out. If the boat proceeds at its maximum speed of 20.2km/h relative to the water, what heading should the boat take?

To solve this problem, we need to determine the heading the boat should take to rescue the child. The situation involves a combination of the boat's velocity relative to the water and the current of the river carrying the child downstream. We will resolve the velocity of the boat into components and find the appropriate angle (heading) for the boat to intercept the child.

Here are the steps to solve the problem:

### Known Data:
- Speed of the current: \( v_c = 2.25 \, \text{km/h} \)
- Distance from the shore: \( y = 0.645 \, \text{km} \)
- Distance upstream from the boat landing: \( x = 0.775 \, \text{km} \)
- Speed of the boat relative to the water: \( v_b = 20.2 \, \text{km/h} \)

### Coordinate System:
We assume the boat is at the origin of a coordinate system. The child is at a position relative to the boat, \( (x, y) = (-0.775, 0.645) \).

### Step 1: Components of the Boat's Velocity
The boat's velocity has two components:
- A horizontal component \( v_{bx} \) to overcome the current and move towards the child in the upstream direction.
- A vertical component \( v_{by} \) to move directly towards the shore.

Let the boat's heading angle with respect to the horizontal axis be \( \theta \). The components of the boat's velocity are:
- \( v_{bx} = v_b \cos(\theta) \)
- \( v_{by} = v_b \sin(\theta) \)

The total velocity of the boat relative to the ground must account for the current, so the effective horizontal velocity is:
\[ v_{\text{effective horizontal}} = v_b \cos(\theta) - v_c \]

### Step 2: Time of Travel
The time it takes for the boat to reach the child can be written in terms of the vertical component of the velocity:
\[ t = \frac{y}{v_{by}} = \frac{0.645}{v_b \sin(\theta)} \]

Since the boat must travel a horizontal distance \( x \), the total horizontal distance traveled by the boat in that time is:
\[ x = (v_b \cos(\theta) - v_c) t \]

Substitute the time from the vertical motion into the horizontal equation:
\[ 0.775 = \left(v_b \cos(\theta) - v_c\right) \frac{0.645}{v_b \sin(\theta)} \]

### Step 3: Solve for \( \theta \)
Now, we can solve this equation for \( \theta \):
\[
0.775 = \left(20.2 \cos(\theta) - 2.25\right) \frac{0.645}{20.2 \sin(\theta)}
\]

This equation can be solved numerically to find the angle \( \theta \), which will give the heading that the boat should take.

I'll calculate this value now.The boat should take a heading of approximately \( 35.7^\circ \) upstream from the horizontal axis to successfully intercept the child.

Would you like further details on how this result was derived, or do you have any questions? Here are some related questions that expand on this problem:

1. How does the speed of the current affect the required heading?
2. What would happen if the boat's speed were slower than the current?
3. How long will it take the boat to reach the child?
4. What happens if the child is farther downstream when the boat sets out?
5. What if the boat starts closer to the shore but farther downstream?

**Tip:** When solving problems involving vectors, breaking the motion into components (horizontal and vertical) is a powerful strategy.

Solve a real-world physics problem involving a rescue boat navigating a river with a current. Learn how to calculate the boat's heading using vector decomposition and trigonometry. The boat must intercept a child being carried downstream by solving for the angle of travel. This problem includes the speed of the current, distance to the shore, and boat speed.

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